BHS 120.1 – Optics 1
Notetaker: Kyle Reuter
Date: 10/02/12, 2nd Hour
Page 1
Sample Problem
-
Calculate the power of a convex surface, made of crown glass surrounded by air, when
the radius of curvature is 20cm.
o Given: n=1.00; n’= 1.52; r = +0.20 m; P = ?
 Going in the direction of light so r is positive
 Formula:
 P = (n’-n)/r (in meters)
o P = (1.52-1.00)/+0.20m = +2.60D
 Note: Converging systems are designated by a plus sign
Clicker Question:
- What would happen if the same convex surface made of ophthalmic crown glass with
radius of curvature of 20cm were immersed in water?
o The refractive power will ____________.
 Increase
 Decrease
 Remain unchanged
- Solution:
o Given: n = 1.33; n’ = 1.52; r = 0.20m
o P = (n’-n)/r  P = (1.52-1.33)/+0.20m = + 0.95D
Clicker Question #2:
- Given the same radius of curvature, which of the following would result in the lowest
refractive power for an interface?
 n = 1.00 and n’ = 1.586
 n = 1.33 and n’ = 1.586
 n = 1.498 and n’ = 1.586
o The lowest refractive power will be the answer with the smallest difference
between n and n’
o Higher values for P can be obtained by:
 A larger difference between n and n’
 Make the rare medium rarer; make the dense medium denser
 Have a more curved surface (shorter radius of curvature)
Cardinal Points for Refracting Surfaces
- PPF: Posterior Principle Focus
o The point on the optic axis where parallel light is refracted by the spherical
refracting surface
o Where ever the light focuses/appears to focus is the PPF
o Light must start at infinity
 Object MUST be at infinity for the focal point to be at PPF
o The image point conjugate to optical infinity
 Optical infinity and PPF are conjugate points
o Known as the secondary focal point (F’ or F2)
BHS 120.1 – Optics 1
Notetaker: Kyle Reuter
Date: 10/02/12, 2nd Hour
Page 2
o
-
If going from a rare to a dense medium, we known the ray will move towards
the normal
 In this case, the ray also moves towards the optic axis
 Movement towards the optic axis tells us that it is converging,
moving towards the normal tells us that it’s moving from a rare
to a dense medium
 Because light started from infinity, we can say that the point where they
come together is the PPF
 If light hadn’t originated from infinity, it would just be a focal
point and NOT the PPF
o APF and PPF are properties of the surface (or interface)
o Images only form at PPF if the object is at infinity
 Images do NOT always form at the PPF
Consider the PPF that forms from a divergent surface:
o
o
-
Light enters from infinity
Goes from rare to dense medium
 Light moves towards the normal but away from the optic axis
o To find PPF, take diverging light and extrapolate the rays back
 Where extrapolated rays meet marks the PPF
o Forms a virtual image at PPF because the image is formed in front of the
interface
APF (Anterior Principle Focus)
o That point on the optic axis when light is refracted it is rendered parallel
thereafter
o Light that originates from the APF of the lens, after getting refracted by the lens,
travels as parallel rays of light
BHS 120.1 – Optics 1
Notetaker: Kyle Reuter
Date: 10/02/12, 2nd Hour
Page 3

o
-
When determining APF and PPF, ask yourself, “Is the direction of light entering parallel
to the optic axis from infinity, or is it leaving parallel to the optic axis toward infinity?”
o If light is entering parallel to the optic axis from infinity = PPF
o If light leaving the interface parallel to the optic axis toward infinity = APF
o
o
o
-
-
i.e. With PPF, light starts at infinity (parallel to optic axis) and forms a
image point along the optic axis; With APF, light starts at a point along
the optic axis and forms rays that run parallel (towards infinity)
Called an object point because it is conjugate to optical infinity
f  Anterior principle focal length = v to APF
f’  Posterior principle focal length = v to PPF
r  Radius of the surface = v to c
Secondary Focal Length (f’ or f2)
o The image distance when the object is at optical infinity
o When calculating the secondary focal length, you must use n’
o To find F’:
 F’ = P + F
 P = refractive power
 F’ = P (with incident plane waves, F = 0)
 P = n’/F’
Primary Focal Length (f or f1)
o The object distance when the image is formed at optical infinity
o Distance from the interface to the APF
o When solving for f, you must use n (primary index of refraction)
o To find f:
 F’ = P + F
 F = -P
BHS 120.1 – Optics 1
Notetaker: Kyle Reuter
Date: 10/02/12, 2nd Hour
Page 4

-
Negative sign is caused because you must reverse the direction
of light to obtain APF
 P = -n/f
We now have two equations: P = n’/f’ and P = -n/f
 Essentially, we have a ratio:
 n’/f’ = -n/f
 Shows that in magnitude, the ratio of the focal lengths equals the ratio of
the refractive indices
 The larger the difference in the refractive indices, the larger the
difference in focal lengths
Sample Problem
- For an air-crown glass interface with radius of curvature of -5.00 cm, calculate the
refractive power and the focal lengths. Is this refracting surface convergent or divergent?
o Given: n = 1.00; n’ = 1.52; r = -5.00cm
 Going rare to dense (n’>n) and negative r
 Gives divergent surface
o Power of refracting surface:
 P = (n’-n)/r
 (1.52-1.00)/-0.05m = -10.4 D
- To calculate f’ (secondary focal length): P = n’/f’
o (1.52)/-10.4D = -0.146 m
-
To calculate f (primary focal length): P = -n/f
o -1.00/-10.4D = 0.096 m
BHS 120.1 – Optics 1
Notetaker: Kyle Reuter
Date: 10/02/12, 2nd Hour
Page 5